A rectangular box has a volume of $4320$ cubic inches and a surface area of $1704$ square inches. The sum of the lengths of its $12$ edges is $208$ inches. What would be the volume of the box, in cubic inches, if its length, width and height were each increased by one inch
To find the volume of the box if its length, width, and height were each increased by one inch, we can first find the dimensions of the original box, and then add one to each dimension to calculate the volume of the new box.
Let's assume the length, width, and height of the original box are $l$, $w$, and $h$ respectively.
From the given information, we have:
Volume of the original box: $lwh = 4320$
Surface area of the original box: $2lw + 2lh + 2wh = 1704$
Sum of the lengths of the edges of the original box: $4l + 4w + 4h = 208$
We can use these equations to find the values of $l$, $w$, and $h$.
From the surface area equation, we can solve for one variable in terms of the other two. Let's solve for $h$:
$2lw + 2lh + 2wh = 1704$
Divide by 2:
$lw + lh + wh = 852$
Rearrange terms:
$h(l + w) = 852 - lw$
Now, we have an equation in terms of $l$ and $w$. Let's call this equation (1).
Next, we can use the sum of the lengths of the edges equation to find another relationship between $l$, $w$, and $h$:
$4l + 4w + 4h = 208$
Divide by 4:
$l + w + h = 52$
Now, we have an equation in terms of $l$, $w$, and $h$. Let's call this equation (2).
At this point, we have two equations, equation (1) and equation (2), that relate $l$, $w$, and $h$. We can solve these equations simultaneously to find the values of $l$, $w$, and $h$.
By substitution, we can substitute the value of $h$ from equation (2) into equation (1), replacing $h$ with $(52 - l - w)$:
$(52 - l - w)(l + w) = 852 - lw$
Expand and simplify the equation:
$52l + 52w - l^2 - w^2 = 852 - lw$
Rearrange terms:
$l^2 + w^2 + lw - 52l - 52w = 852$
Combine like terms:
$l^2 - lw + w^2 - 52l - 52w + 852 = 0$
Now, this equation is a quadratic equation in terms of $l$ and $w$. Let's call this equation (3).
To solve equation (3), we can use the quadratic formula:
$l = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
In this case, $a = 1$, $b = -52$, and $c = 852$. Plugging in these values, we have:
$l = \dfrac{-(-52) \pm \sqrt{(-52)^2 - 4(1)(852)}}{2(1)}$
Simplifying:
$l = \dfrac{52 \pm \sqrt{2704 - 3408}}{2}$
$l = \dfrac{52 \pm \sqrt{-1364}}{2}$
Since the discriminant is negative, there are no real solutions for $l$, and therefore, no real solutions for $w$ and $h$ as well. This means that the given dimensions are not possible for a rectangular box with the given volume, surface area, and sum of lengths of the edges.
Hence, we cannot calculate the volume of the box if its length, width, and height were each increased by one inch, as there is no valid solution for the original box dimensions.